标签：return least line 复杂 wrap ret solution min nta
Find the contiguous subarray within an array (containing at least one number) which has the largest product.
The product of the largest subsequence of the product, less than 2147483647
Example 1:
Input:[2,3,-2,4]
Output:6
Example 2:
Input:[-1,2,4,1] Output:8
思路：动态规划
public class Solution {
/**
* @param nums: An array of integers
* @return: An integer
*/
public int maxProduct(int[] nums) {
int[] max = new int[nums.length];
int[] min = new int[nums.length];
min[0] = max[0] = nums[0];
int result = nums[0];
for (int i = 1; i < nums.length; i++) {
min[i] = max[i] = nums[i];
if (nums[i] > 0) {
max[i] = Math.max(max[i], max[i - 1] * nums[i]);
min[i] = Math.min(min[i], min[i - 1] * nums[i]);
} else if (nums[i] < 0) {
max[i] = Math.max(max[i], min[i - 1] * nums[i]);
min[i] = Math.min(min[i], max[i - 1] * nums[i]);
}
result = Math.max(result, max[i]);
}
return result;
}
}
O(1)空间复杂度
public class Solution {
/**
* @param nums: an array of integers
* @return: an integer
*/
public int maxProduct(int[] nums) {
// write your code here
if (nums == null || nums.length == 0) {
return 0;
}
int minPre = nums[0], maxPre = nums[0];
int max = nums[0], min = nums[0];
int res = nums[0];
for (int i = 1; i < nums.length; i ++) {
max = Math.max(nums[i], Math.max(maxPre * nums[i], minPre * nums[i]));
min = Math.min(nums[i], Math.min(maxPre * nums[i], minPre * nums[i]));
res = Math.max(res, max);
maxPre = max;
minPre = min;
}
return res;
}
}
标签：return least line 复杂 wrap ret solution min nta
原文地址：https://www.cnblogs.com/FLAGyuri/p/12078254.html
